There are 100 people standing in line facing only one direction. They wear either a BLACK HAT or WHITE HAT but they don't know the color of the hat they are wearing. They can see all the hats directly in front of them. For example, person 97 can see and count all the hats from person 1 to 96, but has no clue on the color of his/her own hat as well as those of persons 98-100.

Now, a proctor will ask each person "what is the color of your hat?" starting from the 100th person up to the 1st (100th, 99th, 98th...1st). A person who says the wrong anwer will be eliminated.

Other assumptions:
* each person hears all the previous answers in front of him/her.
* there is no restriction on the # of black/white hats.
there can be 100W - 0B, or 50W - 50B, or 36W- 64B.
* assume that once the questioning is started, the players will always use the startegy and will never make a hum/carelessness error.

Devise a strategy that will lead to the least elimination as possible.

My guess is the person should choose the color the same as the person in front of him/her. But i don't know how to back it up with probabilities. SO please please help me..

2. This seems like a fun question. I'm not 100% certain but here's my idea. According to your method. Only the first person is 100% safe. If the hats are arranged as BWBWBWBW..... then the first 99 will all be eliminated. But if the 1ooth person says the color of the of the 1st, the 99th says the color of the 2nd, 98th says 3rd, and so on, the first 50 is 100% safe. this is my 2 cents.

3. Assuming random hat coloring, each person has a 50% chance at safety. The only way we can change that is by making their hat color known.

The starting person (person 100) has a 50% chance. We can't change this. So we might as well have person100 give person99 a 100% chance at safety by announcing his hat color.

So far we have:
person100: 50%
person99: 100%

But now person98 has got no information about the color of his hat, so it seems that for every two people, one is completely safe while the other is at a 50% chance of elimination.

Let's cheat! I'll be person99. My answer will be given to me by person100. I want to continue the chain and let person98 know what to say while still saying my own hat color. So, I will announce my hat color in a brittish accent if the person ahead of me has a matching hat color, and with an american accent if the person ahead of me has a nonmatching hat color.

Example:

100 has black, 99 has black, 98 has white, 97 has white...

100 guesses "black" and has a 50% chance at safety (luck based).
99 says "black" (because that's what 100 said) with an american accent (because 98 does not match his hat color).
98 says "white" (because he must not match 99) with a brittish accent (because 97 does match his own hat color).

This method will guarantee 99 safe players and 1 player with a 50% chance at elimination. Basically, we added an extra dimension to our span of answers to remove one number in the denominator of our probabilities.

4. But isn't that cheating?

5. i like how you think wytiaz.. it is said that they should devise a strategy..so it isn't necessarily cheating..(i think?) haha thanks for the replies everyone!